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Suppose we have two projective measurements with elements $E_i$, $i=1...m$, and $F_j$, $j=1...n$. So we know $F_j^2=F_j$ and $E_i^2=E_i$ and $\sum_i E_i = \sum_j F_j = \mathbb{I}$. Then it is easy to see that the elements $E_i F_j E_i$ form a POVM, i.e. $E_i F_j E_i\geq 0$ and $\sum_{ij} E_i F_j E_i = \mathbb{I}$. I could imagine that such a composition of measurements appears in several different contexts and I wonder if there is a name for it? Also if it is easy to implement the measurements $E_i$ and $F_j$, then is there a way to do the combined measurement (e.g. some circuit other than just the general constructions that work for every POVM)?

Edit: The post-measurement state for the outcome $(i,j)$ should be $$|\psi'\rangle \propto \sqrt{E_i F_j E_i}|\psi\rangle$$ up to some unitary degree of freedom, see this wikipedia article.

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  • $\begingroup$ Isn't this just equivalent to measuring $E$ first and then measuring $F$, when we assume the standard projective measurement update rule? If you wanted to extend it to POVM's you could just change it to $\sqrt{E_i}F_j\sqrt{E_i}$ and you'd have the same interpretation as a sequential measurement provided you use the Lüder's update rule. $\endgroup$
    – Rammus
    Mar 2, 2022 at 14:04
  • $\begingroup$ @Rammus I don't think it is the same, because the two measurements might not commute. $\endgroup$
    – M. Stern
    Mar 2, 2022 at 14:08
  • $\begingroup$ I might be being very stupid but does that even matter here? I've defined an ordering for the measurements, I'm merely asking what's the probability I get outcome $i$ for measurement 1 and outcome $j$ for measurement 2. This should then be given by $$ \langle \psi| E_i F_j E_i |\psi\rangle. $$ If you wanted the probability for if you measure $F$ first and then $E$ you could measure $F_jE_iF_j$. I'm not asking that the properties specified by $E$ and $F$ are jointly measureable, which indeed would be an issue if the measurements do not commute. $\endgroup$
    – Rammus
    Mar 2, 2022 at 14:13
  • $\begingroup$ @Rammus Hm, yes, that measurement gives the same probabilities for the outcomes, but it does not give the same post-measurement state? $\endgroup$
    – M. Stern
    Mar 2, 2022 at 14:20
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    $\begingroup$ No I mean adjoint (Hermitian conjugate / dagger). Sure, if you want to define the state update in another way then it no longer works. But my comments were to tell you that there is a sense in which it represents a sequential measurement. $\endgroup$
    – Rammus
    Mar 2, 2022 at 15:08

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I'd say that the POVM with elements $\mu\equiv \{E_i F_j E_i\}_{i,j}$, where $\mu_E\equiv \{E_i\}_i$ and $\mu_F\equiv\{F_j\}_j$ are both projective POVMs, can be understood as a "fine-graining" of the POVM $\mu_E$.

More specifically, $E_i F_j E_i$ is the projector onto the linear space $\operatorname{supp}(E_i)\cap\operatorname{supp}(F_j)$, where $\operatorname{supp}(E_i)\equiv I-\operatorname{ker}(E_i)$ is the support of the projector $E_i$. To see this, just observe thta $E_i F_j E_i$ acts as the identity on any $x\in\operatorname{supp}(E_i)\cap\operatorname{supp}(F_j)$, and sends to zero any vector not in this subspace.

As a simple example, consider a three-dimensional space, and POVMs $$\mu_E\equiv \{|+\rangle\!\langle+|, I_3 - |+\rangle\!\langle+|\}, \qquad \mu_F \equiv \{|0\rangle\!\langle0|, |1\rangle\!\langle1|, |2\rangle\!\langle2|\},$$ with $|+\rangle\equiv\frac1{\sqrt2}(|0\rangle+|1\rangle)$. Then, computing $E_i F_j E_i$ for all $i,j$, and summing together identical projections arising from this, we find the associated POVM to be $$\mu = \{|+\rangle\!\langle+|, |-\rangle\!\langle-|, |2\rangle\!\langle2|\}.$$

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  • $\begingroup$ That's an interesting observation and "fine-graining" seems to be a good term for it. $\endgroup$
    – M. Stern
    Mar 3, 2022 at 8:48

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